Q. If $\sin^{- 1}{x - \cos^{- 1}{x = \frac{\pi}{6},}}$ then $x$ is

A) $\frac{1}{2}$
B) $\frac{\sqrt{3}}{2}$
C) $- \frac{1}{2}$
D) None of these
Explanation

(b)
$\sin^{- 1}{x - \cos^{- 1}{x = \frac{\pi}{6}}}$
$\Rightarrow \left( \frac{\pi}{2} - \cos^{- 1}x \right) - \cos^{- 1}{x = \frac{\pi}{6}}$
$\Rightarrow 2\cos^{- 1}{x = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}}$
$\Rightarrow \cos^{- 1}{x = \frac{\pi}{6} \Rightarrow x = \frac{\sqrt{3}}{2}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.